Re: [R] Survreg Weibull lambda and pYes, that is correct! Best, Ales Ziberna ----- Original Message ----- From: Stephen To: Ales Ziberna ; r-help@stat.math.ethz.ch Sent: Thursday, November 24, 2005 4:17 PM Subject: RE: [R] Survreg Weibull lambda and p
Hi Ales, Sorry Mis-read your posting - I blindly saw 'no effect' and was thrown - so its like a dummy regression you loose 1 category I can conclude that p of randgrp is 1.47e-04 or 0.000147 which is a significant difference between the groups? Sorry this is the first time I have done this. Thanks S ------------------------------------------------------------------------------ From: Ales Ziberna [mailto:[EMAIL PROTECTED] Sent: Thu 24/11/2005 14:12 To: Stephen; r-help@stat.math.ethz.ch Subject: Re: [R] Survreg Weibull lambda and p Firstly, I assume that your variable is a numeric one. For seperat values p and lambda for diferent categories, you should convert it to factor. However, this has no effect in your case, since you have only 2 categories. You can have only one p and lambda for a variable with only 2 values. The model can only evaluate the diference, which is what you got, assuming that your groups are coded in such a way, that the difference between the codes is 1. Best, Ales Ziberna ----- Original Message ----- From: "Stephen" <[EMAIL PROTECTED]> To: <r-help@stat.math.ethz.ch> Sent: Thursday, November 24, 2005 10:13 AM Subject: [R] Survreg Weibull lambda and p > Hi All, > > > > I have conducted the following survival analysis which appears to be OK > > (thanks BRipley for solving my earlier problem). > > > >> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, > dist="weibull", scale = 1) > >> summary(surv.mod1) > > > > Call: > > survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, > > dist = "weibull", scale = 1) > > Value Std. Error z p > > (Intercept) 7.36 0.259 28.42 1.27e-177 > > randgrpc -0.59 0.156 -3.80 1.47e-04 > > > > Scale fixed at 1 > > > > Weibull distribution > > Loglik(model)= -1268.6 Loglik(intercept only)= -1276 > > Chisq= 14.72 on 1 degrees of freedom, p= 0.00012 > > Number of Newton-Raphson Iterations: 5 > > n= 400 > > > >> version > > _ > > platform i386-pc-mingw32 > > arch i386 > > os mingw32 > > system i386, mingw32 > > status > > major 2 > > minor 1.1 > > year 2005 > > month 06 > > day 20 > > language R > >> > > I emailed this output to a colleague and received an email requesting > for the 2 groups (randgrpc in the code) > > the 'Weibull lambda and p values' in the analysis. I checked the > mailings for direction but to no avail. > > Could someone please provide direction as how to extract the Weibull > "lambda" and "p" for both randgrpc values? > > Many thanks > > S. > > > ???? ?"? ???? ???? > http://mail.nana.co.il > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html ???? ?"? ???? ???? http://mail.nana.co.il [[alternative HTML version deleted]] ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
